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Kepler's 3rd Law and the Doppler Effect 
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#1
Oct905, 07:01 PM

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Question:
Imagine a space probe has been placed in a circular orbit about a distant planet. The probe emits a continuous radio signal with a wavelength of 8 m. You measure the signal from earth, and find it to have a wavelength that varies regularly between 7.99943 m and 8.00057 m, with a period of 4.5 hours. Assuming that you are in the plane of the probe's orbit, and that you are not moving, calculate the mass of the planet. This is what I have done... By substituting into different equations, I end up with the equation: a = λ / 2pi Using Kepler's 3rd Law: P^2 = a^3 / mt I end up with : mt= (λ / 2pi)^3(1 / P)^2 However, I do not know why I am given 3 different values for wavelength, should this be applied in to the answer or not? 


#2
Oct905, 07:12 PM

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Try instead using: [tex]\frac{\Delta \lambda}{\lambda}=\frac{v}{c}[/tex] and [tex]v_c=\sqrt{\frac{GM}{a}}[/tex] 


#3
Oct905, 07:16 PM

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I just did what my teaching assistant told me to do. I don't think it's correct either.
I'm assuming that "c" is the speed of light. What is the M? Mass of the planet? 


#4
Oct905, 07:17 PM

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Kepler's 3rd Law and the Doppler Effect
And what can I do with that equation for v?



#5
Oct905, 07:19 PM

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#6
Oct905, 07:23 PM

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We haven't learned that much information in order to use those equations. The most we have learned is Kepler's 3rd Law and the Gravitational force equation.
F=GMm/ R^2 


#7
Oct905, 07:37 PM

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#8
Oct905, 07:43 PM

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Okay, thank you anyway.



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